Question

$\text{Standard enthalpy of vaporisation for CCl}_4 \text{ is 30.5 kJ mol}^{-1}. \text{ Heat required for vaporisation of } 284 \, \text{g of CCl}_4 \text{ at constant temperature is } \_\_\_\_ \, \text{kJ.} \\ \text{(Given molar mass in g mol}^{-1}\text{; C = 12, Cl = 35.5)}$

Answer 1

Correct Answer: 56

The standard enthalpy of vaporisation \((\Delta H_v)\) for \(\text{CCl}_4\) is given as \(30.5 \, \text{kJ mol}^{-1}\). We need to find the heat required to vaporize \(284 \, \text{g}\) of \(\text{CCl}_4\).
First, calculate the molar mass of \(\text{CCl}_4\). The formula is \(\text{CCl}_4\), which consists of 1 carbon (C) atom and 4 chlorine (Cl) atoms:

• \(\text{Molar mass of C} = 12 \, \text{g mol}^{-1}\)
• \(\text{Molar mass of Cl} = 35.5 \, \text{g mol}^{-1}\)
• Molar mass of \(\text{CCl}_4 = 1 \times 12 + 4 \times 35.5 = 12 + 142 = 154 \, \text{g mol}^{-1}\)

Next, determine the moles of \(\text{CCl}_4\) in \(284 \, \text{g}\):

\[\text{Moles of CCl}_4 = \frac{284 \, \text{g}}{154 \, \text{g mol}^{-1}} = \frac{284}{154} = 1.844 \, \text{mol}\]Now calculate the heat required for the vaporization using the formula:

\[\text{Heat required} = \text{moles} \times \Delta H_v = 1.844 \, \text{mol} \times 30.5 \, \text{kJ mol}^{-1} = 56.222 \, \text{kJ}\]

Verify this value against the given range (56,56): Since the value rounds to \(56 \, \text{kJ}\).
 

Answer 2

\(\Delta H_{\text{vap}}^\circ \, \text{for } \, \text{CCl}_4 = 30.5 \, \text{kJ/mol}\)

Mass of \( \text{CCl}_4 = 284 \, \text{g} \)  
Molar mass of \( \text{CCl}_4 \):  
\(= 154 \, \text{g/mol}\)

Moles of \( \text{CCl}_4 \):  
\(= \frac{284}{154} = 1.844 \, \text{mol}\)

\(\Delta H_{\text{vap}}^\circ \, \text{for 1 mole} = 30.5 \, \text{kJ/mol}\)

\(\Delta H_{\text{vap}}^\circ \, \text{for 1.844 moles} = 30.5 \times 1.844\)

\(= 56.242 \, \text{kJ}\)

The Correct Answer is:= \(56.242 \, \text{kJ}\)